$$\sum_{i=1}^{n} i^{3} = \frac 14(n^4 - 3n^2 + 2n^2 + 4)$$
This is what I have so far : proof by induction
Base Case : for n = 1 $$1 = \frac 14(1^4-3(1)^3+2(1)^2 + 4)$$ $$1 = 1$$
Base case is true
Induction step: Assume the above is true for n = k and now show it for n = k+1
$$\frac 14(k^4 - 3k^2 + 2k^2 + 4) + (k+1)^3 = \frac 14((k+1)^4 - 3(k+1)^2 + 2(k+1)^2 + 4)$$
$$\frac 14(k^4-3k^3+2k+4) + \frac 44(k^3+3k^2+3k+1) = \frac 14(k^4 + k^3-k^2-3k + 4)$$ $$\frac 14(k^4+k^3+12k^2+14k+8) = \frac 14(k^4 + k^3-k^2-3k + 4)$$
which is obviously not the same thing. I don't know what am I doing wrong here. Can this be prove or there is a counter example for it.
I Got it thanks. This is false claim because if we let n =2 we can give a counter example
$$2^3 = (2^4-3(2)^3+2(2)^2 + 4)/4$$ $$8 = (16-24+8 + 4)/4$$ $$8 != 1 $$
Hense its false